Computer System Architecture - Central Processing Unit
- Option : D
- Explanation : If number is even then LSB bit should be 0.
So total 31 1’s for an unsigned EVEN integer.
And 31 left shifts are needed to determine number of 1’s.
- Option : D
- Explanation : Given: A sequence of micro operations
MBR ⇽ PC
MAR ⇽ X
PC ⇽ Y
Memory ⇽ MBR
To find: What these micro operation represents?
Analysis: On conver ting these operations to normal english
1. First micro operation stores the value of PC into memory Base register (MBR)
2. Second micro operation stores value of X into Memory Address register (MAR)
3. Third micro operation stores value of Y into PC
4. Fourth micro operation stores value of MBR to memory.
So before execution of these instructions PC holds the value of next instruction to be executed. We first stores the value of PC to MBR and then through MBR to memory i.e. we are saving the value of PC in memory and then load PC with a new value. This can be done only in two types of operations conditional branch and interrupt service. As we are not checking here for any conditions so it is interrupt service.
Consider the following program segment for a hypothetical CPU having three user registers R1, R2 and R3.
- Option : B
- Explanation : If memory is word addressable then 1st instruction require two words space (1000 – 1001) 2nd instruction requires only 1 word space ( 1002) and instruction 3rd requires only 1 word space (1003) hence 3rd is add instruction and interrupt occurs during that so after executing that instruction CPU handles the interrupt so return address is 1004
